Question

The vertex of a parabola is (-1.5, -12.5), and its y-intercept is (0, -8).
The x-intercepts of the parabola are
a. (-2,0)
b. (-4,0)
c. (2,0)
d. (4,0)
and
a. (-1,0)
b. (-2,0)
c. (1,0)
d. (2,0)

Answer 1

Vertex for is \(y=a(x-h)^2+k\) where \((h,k)\) is the vertex.

So \(y=a(x+1.5)^2-12.5\)

When we put in \(0\) we get \(-8\) so we use this to solve for \(a\).

\(-8=a(0+1.5)-12.5\implies 4.5=1.5a\implies a = \dfrac{4.5}{1.5}=3\)

So \(y=3(x+1.5)^2-12.5\)

Now set equal to \(0\) and solve.

\(3(x+1.5)^2-12.5=0\implies x+1.5=\pm\sqrt{\dfrac{12.5}{3}} \implies x=\pm\sqrt{\dfrac{12.5}{3}}-1.5\)